Summer High School 2009
Aaron Bertram
8. Primitive Elements and Quadratic Reciprocity (Part I).
Fermat’s Little Theorem tells us that for all nonzero remainders m,
mp−1 ≡ 1 mod p.
Thus every number from 1 to p − 1 is a root of xp−1 − 1 mod p, so
xp−1 − 1 ≡ (x − 1)(x − 2)(x − 3) · · · (x − (p − 1)) mod p
and since the constant term on the right is (−1)p−1 (p − 1)!, we get
Wilson’s Theorem: If p is any prime, then:
(p − 1)! ≡ −1 mod p
(when p is odd, p − 1 is even whereas mod 2 we have −1 ≡ +1).
Question: What is (n − 1)! mod n when n is composite?
Definition 8.1. The order of any nonzero m mod p is the smallest
positive value of d such that:
md ≡ 1 mod p
The order is, of course, always ≤ p − 1, but in fact it divides p − 1.
That’s because of Euclid’s Algorithm. Suppose mk ≡ 1 mod p, and let
d = GCD(k, p − 1). Then we can solve d = ak + b(p − 1), and:
md ≡ mak+b(p−1) ≡ mak mb(p−1) ≡ (mk )a (mp−1 )b ≡ 1 mod p
giving us a smaller power of m that divides p − 1.
Definition 8.2. m mod p is primitive if its order is exactly p − 1. In
this case, the powers of m:
m, m2 , m3 , m4 , . . . , mp−2 , mp−1 ≡ 1 mod p
are all the remainders mod p (in some other order) because if ma ≡ mb
for some a < b ≤ p − 1, then mb−a ≡ 1, which isn’t allowed.
For example, mod 11:
m ≡ 2, m2 ≡ 4, m3 ≡ 8, m4 ≡ 5, m5 ≡ 10
m6 ≡ 9, m7 ≡ 7, m8 ≡ 3, m9 ≡ 6, m10 ≡ 1
and notice that m5 ≡ 10 ≡ −1 mod 11.
Lemma 8.3 For each d that divides p − 1, there are φ(d) remainders
of order d. In particular, there are φ(p − 1) primitive remainders.
1
2
Proof: If m has order d, then the roots of xd − 1 are exactly the
remainders m, m2 , m3 , . . . , md−1 , md ≡ 1. Consider mk for some k < d.
If GCD(k, d) = e 6= 1, then (mk )(d/e) = (m(k/e) )d ≡ 1 has smaller order.
Thus the only possible remainders of order d are the powers of m that
are relatively prime to d. Thus there are at most φ(d) remainders of
order d for each d that divides p − 1. On the other hand, if you add all
of the values of the φ function for divisors of p − 1 together:
X
φ(d) = p − 1
d|p−1
so there must be exactly φ(d) remainders of each order d, otherwise we
wouldn’t be able to account for all the numbers from 1 to p − 1.
Examples: Mod 11, we consider the divisors of 10 = 11 − 1:
φ(1) + φ(2) + φ(5) + φ(10) = 1 + 1 + 4 + 4 = 10
and the corresponding remainders are:
1 (order 1), 10 (order 2), 4, 5, 9, 3 (order 5), 2, 8, 7, 6 (order 10)
P
Food for thought: Why is it that d|n φ(n) = n for all n?
We next consider: “What are the perfect squares mod p?”
For starters, 1 is obviously always a square, with square roots 1, −1
(which are different from each other as long as p is an odd prime).
The next easiest case is m = −1:
Proposition 8.4. Let p be an odd prime, so p ≡ 1 or p ≡ 3 mod 4.
(a) −1 is not a square mod p if p ≡ 3 mod 4.
(b) −1 is a square mod p if p ≡ 1 mod 4. Moreover, in that case:
p−1
p−1
p−1
! = 1 · 2 · 3···
, and −
!
2
2
2
are the two square roots of −1 modulo p.
Proof: Suppose m is a primitive remainder mod p. Remember that:
m, m2 , m3 , . . . , mp−1 ≡ 1 mod p
runs through all the remainders modulo p. Since each square mod p
has two different square roots (r and −r), it follows that exactly half
of the remainders mod p are squares. But:
m2 , (m2 )2 ≡ m4 , (m3 )2 ≡ m6 , . . . , (m
p−1
2
)2 ≡ mp−1
are all different, and all perfect squares, so they must be all of them!
3
p−1
p−1
Notice that (m 2 )2 ≡ 1, and m 2 6≡ 1 mod p, so it must be −1. If
p−1
2
p ≡ 3 mod 4, then p−1
is
odd,
so
m
is NOT a square, and if p ≡ 1
2
p−1
mod 4, then 2 is even, so it is a perfect square, and
mod
if p ≡ 1p−1
4,
p−1
p−1
then p − 1 ≡ −1, p − 2 ≡ −2, . . . , p − 2 = 2 + 1 ≡ − 2 , so
by Wilson’s Theorem:
2
p−1
p−1
p
−
1
p
−
1
−1 ≡ (p−1)! ≡
!·(−1)( 2 )
!≡
!
mod p
2
2
2
Examples: The first primes ≡ 1 mod 4 are: 5, 13, 17 and 29.
(a) (mod 5) 2! = 2, and 22 ≡ −1 mod 5.
(b) (mod 13) 6! = 720 ≡ 5 mod 13, and 52 = 25 ≡ −1 mod 13.
(c) (mod 17) 8! = 40320 ≡ 13 mod 17, and 132 = 169 ≡ −1 mod 17.
(d) (mod 29) 14! ≡ 12 mod 29, and 122 = 144 ≡ −1 mod 29.
Remark: The nice thing about this Proposition is that it tells us
whether or not −1 is a perfect square without requiring us to find
the square
root. It just so happens that the square root is given by
p−1
!,
but
this is actually quite hard to calculate when p is large.
2
Quadratic Reciprocity will similarly tell us whether any remainder
mod p is a perfect square or not, with a minimal amount of checking
of congruences mod 4 (and mod 8). We will follow one of Gauss’ many
proofs of this, which proceeds in stages.
Lemma 8.5 (Stage 1): Let a mod p be nonzero. Then:
p−1
a( 2 ) ≡ 1 or − 1 mod p
and a is a square if it is 1, and not a square if it is −1.
Proof: Let m be some primitive mod p, and choose k so that a ≡ mk .
p−1
• If k = 2l is even, then a is a square and a( 2 ) ≡ m(p−1)l ≡ 1 mod p.
• If k = 2l + 1 is odd, then a is not a square, and
p−1
p−1
a( 2 ) ≡ m(p−1)l+( 2 ) ≡ −1 mod p. Example: mod 11:
15 ≡ 1 mod 11. Perfect square (duh!).
25 = 32 ≡ −1 mod 11. Not a perfect square.
35 = 243 ≡ 1 mod 11. Perfect square.
45 = 1024 ≡ 1 mod 11. Perfect square (duh!).
55 = 3125 ≡ 1 mod 11. Perfect square.
4
65 ≡ (−5)5 ≡ −1 mod 11. Not a perfect square.
75 ≡ (−4)5 ≡ −1 mod 11. Not a perfect square.
85 ≡ (−3)5 ≡ −1 mod 11. Not a perfect square.
95 ≡ (−2)5 ≡ 1 mod 11. Perfect square (duh!).
105 ≡ (−1)5 ≡ −1 mod 11. Not a perfect square.
This is somewhat nice, but it involves too many calculations. It is,
however, the key ingredient in a further extremely clever Lemma due
to Gauss. For this, we consider the first half of the multiples of a:
p−1
a
a, 2a, 3a, 4a, . . . ,
2
p−1
and we choose remainders that lie between − p−1
and
mod p
2
2
(rather than the usual remainders between 1 and p − 1). Let n be the
number of negative remainders.
Gauss’ Lemma 8.6 (Stage 2): Let a be nonzero mod p, and let n
be defined as above. Then:
• a is a square mod p if n is even, and:
• a is not a square mod p if n is odd.
Proof: Since
2)a,
−a ≡ (p − 1)a, −2a ≡ (p −p−1
. . . it follows
that when
p−1
p−1
a, 2a, . . . , 2 a are brought between − 2 and 2 mod p, then
at most one of each of the following pairs arises:
p−1
p−1
1 or − 1, 2 or − 2, 3 or − 3, . . . ,
or −
2
2
p−1
but since there are exactly 2 of these, it follows that one of each
pair does arise. Multiply all the remainders together:
p−1
p−1
(a)(2a)(3a) · · ·
a ≡ (±1)(±2)(±3) · · · ±
mod p
2
2
and exactly n of the “±”s on the right side is a “−.” As in the proof
! from both sides to get:
of Euler’s formula, we can now cancel p−1
2
a(
p−1
2
) ≡ (−1)n mod p
which, together with Stage 1, proves Stage 2.
Example: (a) Check that 2 is not a square mod 11.
2, 4, 6, 8, 10 become 2, 4, −5, −3, −1
so n = 3 is odd, and 2 is not a square.
5
(b) Check that 3 is a square mod 11.
3, 6, 9, 12, 15 become 3, −5, −2, 1, 4
so n = 2 is even, and 3 is a square.
We can generalize Example (a) in a big way to get the following:
Proposition 8.7. If p is an odd prime, p ≡ 1, 3, 5 or 7 mod 8, and:
(a) 2 is a square mod p if p ≡ 1 or 7 mod 8, and
(b) 2 is not a square mod p if p ≡ 3 or 5 mod 8.
Proof: As in Example (a), consider:
p−1
2, 4, 6, 8, · · · , p − 1 = 2
2
and rememberthat n is the number of these that are larger than p−1
2
minus the number that are less than or equal to p−1
).
(which is p−1
2
2
• If p = 8l + 1, then n = 4l − 2l = 2l is even.
• If p = 8l + 3, then n = (4l + 1) − 2l = 2l + 1 is odd.
• If p = 8l + 5, then n = (4l + 2) − (2l + 1) = (2l + 1) is odd.
• If p = 8l + 7, then n = (4l + 3) − (2l + 1) = (2l + 2) is even.
Examples:
(a) 2 is not a square mod 3.
(b) 2 is not a square mod 5.
(c) 2 ≡ 32 mod 7.
(d) 2 is not a square mod 11 (and 11 ≡ 3 mod 8).
(e) 2 is not a square mod 13 (and 13 ≡ 5 mod 8).
(f) 2 ≡ 62 mod 17 (and 17 ≡ 1 mod 8).
(g) 2 is not a square mod 19 (and 19 ≡ 3 mod 8).
(h) 2 ≡ 52 mod 23 (and 23 ≡ 7 mod 8).
Again, notice that the Proposition tells us whether 2 is a perfect
square or not mod p simply by testing p mod 8.